Question

evaluate the following limits. if needed, enter oo for \\( \infty \\) and -oo for \\( -\infty \\).
(a) \\( \lim _{x \
ightarrow \infty} \frac{\sqrt{11+8 x^{2}}}{8+10 x}= \\)
(b) \\( \lim _{x \
ightarrow-\infty} \frac{\sqrt{11+8 x^{2}}}{8+10 x}= \\)
question help: video message instructor

Explanation:

Step1: Factor out \(x^2\) from the square - root

For \(\lim_{x
ightarrow\infty}\frac{\sqrt{11 + 8x^{2}}}{8+10x}\), we know that \(\sqrt{11 + 8x^{2}}=\sqrt{x^{2}(\frac{11}{x^{2}}+8)}=\vert x\vert\sqrt{\frac{11}{x^{2}}+8}\). When \(x
ightarrow\infty\), \(\vert x\vert=x\). So \(\frac{\sqrt{11 + 8x^{2}}}{8+10x}=\frac{x\sqrt{\frac{11}{x^{2}}+8}}{8 + 10x}\).
Divide both the numerator and denominator by \(x\):

$$
LATEXBLOCK0
$$

Step2: Use the limit property \(\lim_{x

ightarrow\infty}\frac{c}{x^{n}} = 0\) (\(n>0,c\) is a constant)
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{11}{x^{2}}=0\) and \(\lim_{x
ightarrow\infty}\frac{8}{x}=0\). Then \(\lim_{x
ightarrow\infty}\frac{\sqrt{\frac{11}{x^{2}}+8}}{\frac{8}{x}+10}=\frac{\sqrt{0 + 8}}{0+10}=\frac{2\sqrt{2}}{10}=\frac{\sqrt{2}}{5}\).

Step3: Deal with \(x

ightarrow-\infty\) case
For \(\lim_{x
ightarrow-\infty}\frac{\sqrt{11 + 8x^{2}}}{8+10x}\), \(\sqrt{11 + 8x^{2}}=\sqrt{x^{2}(\frac{11}{x^{2}}+8)}=\vert x\vert\sqrt{\frac{11}{x^{2}}+8}\). When \(x
ightarrow-\infty\), \(\vert x\vert=-x\). So \(\frac{\sqrt{11 + 8x^{2}}}{8+10x}=\frac{-x\sqrt{\frac{11}{x^{2}}+8}}{8 + 10x}\).
Divide both the numerator and denominator by \(x\) (\(x<0\), so \(\frac{-x}{x}=- 1\)):

$$
LATEXBLOCK1
$$

Step4: Apply the limit property \(\lim_{x

ightarrow-\infty}\frac{c}{x^{n}} = 0\) (\(n>0,c\) is a constant)
As \(x
ightarrow-\infty\), \(\lim_{x
ightarrow-\infty}\frac{11}{x^{2}}=0\) and \(\lim_{x
ightarrow-\infty}\frac{8}{x}=0\). Then \(\lim_{x
ightarrow-\infty}\frac{-\sqrt{\frac{11}{x^{2}}+8}}{\frac{8}{x}+10}=\frac{-\sqrt{0 + 8}}{0+10}=-\frac{\sqrt{2}}{5}\)

Answer:

(a) \(\frac{\sqrt{2}}{5}\)
(b) \(-\frac{\sqrt{2}}{5}\)